Question

What volume of 0.1 M KMnO₄ solution is needed to completely oxidation of 40 ml of 0.2 M oxalic acid in presence of H₂SO₄?

Answer 1

2KMnO₄ + 3H₂SO₄ →K₂SO₄ + 2MnSO₄ + 3H₂O + 5O

\(\overset{COOH}{\underset{COOH}|}+O\) → 2CO₂ + H₂O]x5

\(\overset {+7} {2KMnO_4}+ 3H_2SO_4 + \overset{+3}{\overset{5\,COOH}{\underset{+3COOH}|}}\)

→ K₂SO₄ + \(\overset {+2}{2MnSO_4} + \overset {+4}{10CO_2} + 8H_2O\)

Change of oxidation state of Mn = 5

∴ Normality of KMnO₄ = molarity x n

= 0.1 x 5

= 0.5 N

Change of Oxidation carbon in oxalic acid = 2

∴ Normality of Oxalic acid = 0.2 M x 2 = 0.4 N

Using Normality equation

(N₁V₁) KMnO₄ = (N₂V₂) oxalic acid

= V₁ x 0.5 N = 0.4N x 40 ml

= V₁ = \(\frac {0.4N x 40ml}{0.5}\)

= V₁ = 32 ml

Hence, 32 ml of 0.1 M KmnO₄ required to complete oxidation of 40 ml of 0.2 oxalic acid.